generated from smyalygames/quartz
36 lines
1.9 KiB
Markdown
36 lines
1.9 KiB
Markdown
---
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lecture: 8
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date: 2025-02-03
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---
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Defines in [[Topological Space|topological spaces]]:
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- [[Connected]]
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- [[Connected Component]]
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# Proposition
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$[0, 1]$ is [[connected]].
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## Proof
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$$[0, 1] = A \cup B$$
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$$A \cap B = \emptyset$$
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Suppose [[Open Sets|open sets]] $A$ and $B$ disconnected it, and say $1 \in B$.
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Then every neighbourhood of $a \equiv \sup{A} \in [0, 1]$ will intersect both $A$ and $B$.
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Which is a contradiction as this is impossible!
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QED.
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# 2.2 Continuity
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Recall: $f: \mathbb{R} \to \mathbb{R}$ is continuous at $x$ if $\lim_{ y \to x }f(y) = f(x)$, and it would be discontinuous if $\lim_{ y \to x } f(y) \neq f(x)$. More precisely, if $\forall \varepsilon \gt 0 \, \exists \delta \gt 0$ such that $|x-y| \lt \delta \implies |f(x) - f(y) < \varepsilon$ or $f(B_{\delta}(x)) \subset B_{\varepsilon}(f(x))$, or $B_{\delta}(x) \subset f^{-1}(B_{\varepsilon}(f(x))) = \{ y \in \mathbb{R} | f(y) \in B_{\varepsilon}(f(x)) \}$.
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## Definition
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The definition is defined at: [[Continuous]].
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# Proposition
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A [[Continuous|continuous]] $f : X \to \mathbb{R}$ with $X$ [[Compact|compact]], it will attain both a maximum and a minimum.
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## Proof
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Note: the [[Continuous|continuous]] image of a [[Compact|compact set]] is [[Compact|compact]] since [[Inverse Images|inverse images]] of an [[Open Cover|open cover]] will again be an [[Open Cover|open cover]]. Then the final step is to use the [[Heine–Borel]] theorem since $f(x)$ is compact in $\mathbb{R}$.
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QED.
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# Something else
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We say $f$ is [[Continuous|continuous]] (at every $x$) if $f^{-1}(A)$ is [[Open Sets|open]] for every open $A \subset Y$.
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Continuous images of [[Connected|connected]] spaces are [[Connected|connected]]. Again because inverse images of [[Open Sets|open subsets]] disconnecting an image would disconnect the domain.
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So homeomorphisms preserve compactness and connectedness. They are [[Topological Invariants|topological invariants]]. |